Q. The equations of the sides $AB,\ BC,\ CA$ of a $\mathrm{\Delta}ABC$ are $x + y = 1,\ 4x - y + 4 = 0$ and$2x + 3y = 6$. Circles are drawn on $AB,\ BC,\ CA$ as diameter. The point of concurrence of the common chord is
A) Centroid of the triangle
B) Orthocenter
C) Circumcentre
D) Incentre
Explanation
(b)
Since, $\angle\ ADB = \angle ADC = 90{^\circ}$, circle on $AB$ and $AC$ as dismeters pass through $D$ and therefore the altitude $AD$ is the common chord. Similarly, the other two common chords are the other two altitudes and hence they concur at the orthocenter
[image:11_image2.emf]
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.